Follow-up Question to Soft-Starter Topic 2

[nightfox1925]

In one of the previous thread237-184975 entitled "Does Soft-Starters Reduce Energy Bills?", Marke mentioned that:

"There are also some who claim that by reducing the start current, you will save energy. This is also not true.
You will use the same amount of energy to start the load, just over a longer period."

Marke (or anybody), can you kindly enlightened me as to why this is not so? I'm also up to the theory that reducing the start current deos not provide any energy savings and you might have a better and more convincing type of expalanation. Thanks

 

[fangas]

Essentially, and very simplistic

Lowering the current will reduce the overall energy. However, it will also reduce the starting time. So, if you reduce current by 50% then take twice as much time to start. It is the same as 100% current in half the time to start.

Ed

 

[jraef]

Keep in mind that current is not energy. Current is not even power. Current is only 1 of 3 components of power and a soft starter is playing with 2 of 3 (really only 1 directly (voltage), but it can also control current indirectly). Energy is Power and Time. If you plotted the Energy required to start a load as a curve using input Power as the Y axis and acceleration Time as the X axis, the DOL curve would be high, steep and short, the soft start curve would be low, gentle, but long. However if you calculate the areas of the curves (the Energy) from stop to full speed, they would be almost exactly the same. The only difference might be an infinitesimal amount of extra I²R losses in the conductors during soft starting because of the increased amount of time at higher current causing the conductors to heat up slightly more and increasing their resistance. Not enough to even measure, let alone worry about. But either way, soft starting does NOT significantly change the energy requirement.

Generally, those who continue to imply that soft starters save energy just by soft starting are in essence showing their ignorance, or worse yet, their total disregard for factual engineering principals. Either way, whatever product they are selling should probably be avoided because of that.


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[LionelHutz]

You could also look at it from a more fundamental perspective. You have to convert electrical energy into angular or rotational kinetic energy to change an object from sitting at rest to rotating at a certain rpm. The amount of energy required is the same no matter how quickly or slowly you accelerate this object. Just the rate of energy transfer changes.

 

[Marke]

Hello nightfox

In order to start a machine, you must a) supply the work output torque across the speed range from 0 - 100% speed and b) provide enough energy to get the machine to full speed.
In order to get to full speed you must supply torque over and above the work torque at all speed such that the surplus energy is accumulated in the inertia of the load.

If we ignore the work component and assume that we have a purely inertial load that is doing nothing and has no losses, then all we have to do is to "charge up the inertia" which is like a capacitor.

The shaft power during start, is proportional to the shaft speed and the shaft torque. At half speed, about half the input power is shaft power, and the other half is slip power (dissipated in the rotor). (there is also stator power loss)
If we halve the start current, we will reduce the torque to one quarter. All losses will also reduce to one quarter.
At a quarter of the torque output, the motor will take four times as long to accelerate the load to full speed, so we have a quarter of the losses for four times as long and so we have the same total losses.

If we now add in some actual work power as well, then the surplus torque will be reduced and the start time further extended, so that the total losses will infact increase with reduced current. - This only becomes a significant issue if the acceleration torque is small relative to the work torque at a range of speeds.

Best regards,

Mark Empson
L M Photonics Ltd

 

[nightfox1925]

Thanak you very much for the clear and patiently delivered explanation and this would greatly help in portraying that saving energy through soft starting is a myth used by sales people.

 

[burnt2x]

nightfox,
My take on this topic is that the "soft-starter" does help in preventing large kVA draw from the system.
If the sales people claim savings, I guess they should be saying savings in terms of first cost as the industry owners will be buying smaller trafos and related equipment. We all agree that "current" itself is not an indication of "power/energy" unless the power factor is unity and the system voltage does not collapse!
When a power system supplies power to a certain user, current draw during start ups are always taken into consideration as well as the delivery point voltage. Otherwise, we will end up with cascading tripping when big blocks of power are drawn from the system and the system collapses.
Hope this addds to the whole understandingof the matter.
respectfully.

 

[waross]

Starting loads may be considered in system design, but in the majority of cases the design may not be changed. There are exceptions.
The heating time constant of a distribution transformer is on the order of 15 minutes. Motor starting is not generally a factor.
Extremely large motors or large motors on soft distribution systems would be exceptions where the system design would be impacted by the motor starting method.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Marke]

I commonly get sales people claiming that the soft starter will reduce the maximum demand penalty, however, in most cases, the maximum demand is averaged over a 15 or 30 minute period, so it is the area under the current curve that is important, not the maximum, and this is higher with reduced voltage starting (half the current for four times as long at half voltage and an inertial load) rather than lower, so that is a myth except where the tariff is based on peak values.

Best regards,

Mark Empson
L M Photonics Ltd

 

[nightfox1925]

Thanks Marke...one follow-up question related to this topic. A soft starter will have SCRs firing and there is this so called current ramping.

Assuming that the soft starter is set to initially reduce the voltage to 75%, then starting current will be reduced 75% as well and torque will be 56.25%.

Is my understanding correct if I say that the current limiter feature of the SS will be set to stop the starting process once the pre-set motor acceleration time is exceeded so as not to exceed the motor thermal limits.

What is the difference between voltage ramp and current ramp associated with SS starting?

 

[LionelHutz]

Voltage ramp - The controls just phase on the SCR's at a predetermined rate to increase the voltage. There is no adjustment to control the current to the motor.

Current ramp - The controls will measure the current and adjust the phase angle of the SCR's to maintain the programmed current.

The current ramp typically is set so that the current is high enough to get the motor to start over a reasonable period. There should be a motor overload that will stop the motor if the thermal limits are being exceeded. The setting of a current ramp mostly depends on the operator to properly pick the values and monitor the motor during the initial programming to ensure it's functioning well.

 

[jraef]

Assuming that the soft starter is set to initially reduce the voltage to 75%, then starting current will be reduced 75% as well and torque will be 56.25%.

Essentially, yes. But keep in mind that the limited values are of the motor's basic torque speed curve, not of FL values. So for instance if you limit the voltage to 75%, the initial starting current will be reduced to 75% of the DOL current, which is already 600% of FLC, therefore your initial current will be 450% (75% of 600%) and your torque will be 56.25% of 160% FLT, or 90% of FLT.

Is my understanding correct if I say that the current limiter feature of the SS will be set to stop the starting process once the pre-set motor acceleration time is exceeded so as not to exceed the motor thermal limits.

First off, realize that Current Ramp and Current Limit are two different things. Speaking then of Current Limit, no, you are misunderstanding that. The Current Limit function of a soft starter will not, in and of itself, monitor what is happening in the motor. It just limits the current to a maximum setting. If that produces insufficient torque to continue accelerating the motor, eventually the Thermal Overload function will be expected to take the starter off-line. Some soft starters also have a maximum Acceleration Time Limit feature which will not wait for the OL function in case the motor does not fully accelerate, but this feature is not universally available unfortunately.

What is the difference between voltage ramp and current ramp associated with SS starting?

I don't have much to add to LionelHutz's response, other than as a gross generality, Current Ramping tends to be a little smoother on some applications, for whatever that's worth. In 99.99% of applications, you will not notice much of a difference.

I do a lot of commissioning for a brand of soft starter that has all available ramp profile methods available in it, so I have tried them all on a few applications. I can barely tell the difference and generally default to Voltage Ramp with a Current Limit.


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[nightfox1925]

"First off, realize that Current Ramp and Current Limit are two different things. Speaking then of Current Limit, no, you are misunderstanding that. The Current Limit function of a soft starter will not, in and of itself, monitor what is happening in the motor. It just limits the current to a maximum setting."

If the current limit is set, then whatever was the reduced voltage at which the current reached this limit, it stops the SCR firing and maintain the current level...not necessarily stops the starting process. is my understanding now correct?

Thank you very much for all the support and enlightenment.

 

[jraef]

If the current limit is set, then whatever was the reduced voltage at which the current reached this limit, it stops the SCR firing and maintain the current level...not necessarily stops the starting process. is my understanding now correct?

Yes. The SCRs of course don't STOP firing, that would be Off! But I know what you meant. The advancement of the firing angle stops so the SCRs continue to provide the same RMS voltage to the motor during Current Limit until either the motor current drops on it's own, meaning it accelerated, or the OL trips (or the Accel timer times out if it's there).


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